Description:
给你一个链表以及一个k,将这个链表从头指针开始每k个翻转一下。
链表元素个数不是k的倍数,最后剩余的不用翻转。
Explanation:
给出链表
1->2->3->4->5
k =
2
, 返回
2->1->4->3->5
k =
3
, 返回
3->2->1->4->5
Solution:
首先判断长度能否翻转。如果能翻转,依次将节点插入头部,直至操作k次。然后递归处理。注意每次返回的结果都是k个长度的子串,每次翻转后都需要拼接。
1 -> 2 -> 3 -> 4 -> 5 pre=null;
| |
c cn
1 -> 2 -> 3 -> 4 -> 5 2 -> 1 -> null
| | |
pre c cn
1 -> 2 -> 3 -> 4 -> 5 3 -> 2 -> 1 -> null
| | | |
pre c cn
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
/**
* @param head a ListNode
* @param k an integer
* @return a ListNode
*/
public ListNode reverseKGroup(ListNode head, int k) {
// Write your code here
if(k == 1) return head;
if(head == null || head.next == null) return head;
ListNode current = head;
ListNode pre = null;
ListNode cnext = null;
ListNode res = null;
if(len(head , k)){
cnext = current.next;
int count = k;
while(count >= 1){
count--;
cnext = current.next;
if(count == 0){
res = current;
}
current.next = pre;
pre = current;
current = cnext;
}
}else{
return head;
}
ListNode last = res;
while(last.next != null){
last = last.next;
}
ListNode next = reverseKGroup(cnext , k);
last.next = next;
return res;
}
public boolean len(ListNode head , int k){
while(head != null){
head = head.next;
k--;
}
return k<=0?true:false;
}
}